Problem: Simplify the following expression and state the condition under which the simplification is valid. $t = \dfrac{8n^3 - 64n^2 + 120n}{6n^2 - 78n + 180}$
First factor out the greatest common factors in the numerator and in the denominator. $ t = \dfrac {8n(n^2 - 8n + 15)} {6(n^2 - 13n + 30)} $ $ t = \dfrac{8n}{6} \cdot \dfrac{n^2 - 8n + 15}{n^2 - 13n + 30} $ Simplify: $ t = \dfrac{4n}{3} \cdot \dfrac{n^2 - 8n + 15}{n^2 - 13n + 30}$ Next factor the numerator and denominator. $ t = \dfrac{4n}{3} \cdot \dfrac{(n - 3)(n - 5)}{(n - 3)(n - 10)}$ Assuming $n \neq 3$ , we can cancel the $n - 3$ $ t = \dfrac{4n}{3} \cdot \dfrac{n - 5}{n - 10}$ Therefore: $ t = \dfrac{ 4n(n - 5)}{ 3(n - 10)}$, $n \neq 3$